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70
Chapter 2 Overview: More Advanced Derivatives
The purpose of this chapter is to build on what you learned last year in precalculus
as well as what you learned in chapter 1 of this book. You remember that one of
the contexts for derivatives last year was finding maxima and minima, as well as
finding the intervals of increasing and decreasing on a function. We will explore
these concepts further in this chapter.
Additionally, we will look at the concept of local linearity. This is basically just
the fact that as you look very closely at a curve, the curve looks pretty much like a
straight line. If you zoom in close enough, the curve and the tangent line end up
looking exactly the same. We can use this fact to approximate values for a
function by using the tangent line – this was one of the early uses for Calculus.
Lastly, we will be looking at the derivatives of inverse functions and higher order
derivatives. Just like we can take the derivative of all the trig functions, we will be
able to take derivatives of all of the trig inverse functions as well. We will also be
looking at the fact that we can take the derivative of a function more than once –
we can keep taking derivative after derivative after derivative if we like.
71
2.1: Higher Order Derivatives
What we have been calling the Derivative is actually the First Derivative. There
can be successive uses of the derivative rules, and they have meanings other than
the slope of the tangent line. In this section, we will explore the process of finding
the higher order derivatives.
Second Derivative--Defn: The derivative of the derivative.
Just as with the First Derivative, there are several symbols for the 2nd Derivative:
Higher Order Derivative Symbols
Liebnitz: d y
dx
2
2 = d squared y, d x squared;d y
dx
3
3;…
n
nd ydx
Function: "f x = f double prime of x; '''f x ; IVf x ;… nf x
Combination: y'' = y double prime
OBJECTIVE
Find higher order derivatives.
Ex 1 d
x x x xdx
2
4 3 22 7 3 2 5
d d dx x x x x x x x
dx dxdx
dx x x
dx
24 3 2 4 3 2
2
3 2
7 3 2 5 7 3 2 5
4 21 6 2
x x 212 42 6
72
Ex 2 Find d y
dx
3
3 if sin y x 3
sin
cos cos
sin sin
cos cos
y x
dyx x
dx
d yx x
dx
d yx x
dx
2
2
3
3
3
3 3 3 3
3 3 3 9 3
9 3 3 27 3
More complicated functions, in particular Composite Functions, have a
complicated process. When the Chain Rule is applied, the answer often becomes a
product or quotient. Therefore, the 2nd Derivative will require the Product or
Quotient Rules as well as, possibly, the Chain Rule again.
Ex 3 xy e23 , find y''.
x x
x x
x x
x
dye x xe
dx
d yx e x e
dx
x e e
e x
2
2
2
2
2 2
2 2
2 2
2
3 3
3 3
3 3
3
6 6
6 6 6
36 6
6 6 1
Ex 4 siny x 3 , find y''
' sin cos
'' sin sin cos sin cos
sin cos sin
y x x
y x x x x x
x x x
2
2
2 2
3
3 6
3 2
73
Ex 5 lnf x x x 2 3 1 , find ’’f x .
’
’’
xf x x
x x x x
x x x xf x
x x
x x x x
x x
x x
x x
2 2
2
22
2 2
22
2
22
1 2 32 3
3 1 3 1
3 1 2 2 3 2 3
3 1
2 6 2 4 12 9
3 1
2 6 11
3 1
Ex 6 g x x 24 1 , find ’’g x .
’
’’
xg x x x
x
x x x x
g x
x
xx
x
x
x x
x
x
122
122
1 12 22 2
122
2122
122
2
2 2
322
322
1 44 1 8
24 1
14 1 4 4 4 1 8
2
24 1
164 1 4
4 1
4 1
4 1 4 16
4 1
4
4 1
74
2.1 Homework Set A
In #1-5, find the first and second derivatives of the function.
1. 5 2( ) 6 7f x x x x 2. 2( ) 1h x x
3. 2
33 1y x 4. ( ) tan3H t t
5. 3 5( ) tg t t e 6. If 23xy e , find ''y .
75
7. If 3siny x , find ''y . 8. If ( ) cosf t t t , find '''(0).f
2.1 Homework Set B
1. Find 'f x , ''f x , and '"f x if ln secf x x .
2. Find dy
d and
2
2
d y
d for 10cot 2 1y
76
3. For the function tan2y x , show that 2
22
8sec 2 tan 2d y
x xdx
.
4. Find "f x for 2ln 3 1f x x x
5. Find the first, second, and third derivative for 2 5 6 xf x x x e
77
6. Find the first, second, and third derivative for tan 3 2f
7. A fourth differentiable function is defined for all real numbers and satisfies
each of the following:
2 5g , ' 2 2g , and " 2 3g
If the function f is given by 1
2k x
f x e g x
, where k is a constant.
a. Find 1f , ' 1f , '' 1f
b. Show that the fourth derivative of f is
14 16 '''' 2k x
k e g x
78
Answers: 2.1 Homework Set A
1. 5 2( ) 6 7f x x x x 2. 2( ) 1h x x
4'( ) 5 12 7f x x x 2
'( )1
xh x
x
3"( ) 20 12f x x
2
322
1
"( )
1
xh x
x
3. 2
33 1y x 4. ( ) tan3H t t
2
133
2
1
dy x
dxx
2'( ) 3sec 3H t t
32
2 433
2 2
1
x xd y
dxx
2 "( ) 18sec 3 tan3 H t t t
5. 3 5( ) tg t t e 6. If 23xy e , find ''y .
2 5'( ) 5 3tg t t e t 23 2'' 6 6 1xy e x
5 2 "( ) 25 30 6tg t te t t
7. If 3siny x , find ''y . 8. If ( ) cosf t t t , find '''(0).f
2 2'' 3sin 2cos siny x x x '''(0) 3f
2.1 Homework Set B
1. Find 'f x , ''f x , and '"f x if ln secf x x .
' tanf x x
2'' secf x x
2''' 2sec tanf x x x
79
2. Find dy
d and
2
2
d y
d for 10cot 2 1y
220csc 2 1d
dy
2
2
280csc 2 1 cot 2 1
d
d y
3. For the function tan2y x , show that 2
22
8sec 2 tan 2d y
x xdx
.
2
2
2
22
2
2sec 2
4sec2 sec2 tan 2 2
8sec 2 tan 2
dy
dx
d y
dx
d y
dx
x
x x x
x x
4. Find "f x for 2ln 3 1f x x x
2
22
2 6 11''
3 1
x xf x
x x
5. Find the first, second, and third derivative for 2 5 6 xf x x x e
' 2 5 xf x x e
'' 2 xf x e
''' xf x e
6. Find the first, second, and third derivative for tan 3 2f
2' 3sec 3 2f
2'' 18sec 3 2 tan 3 2f
2 2 2''' 54sec 3 2 sec 3 2 2tan 3 2f
80
7. A fourth differentiable function is defined for all real numbers and satisfies
each of the following:
2 5g , ' 2 2g , and " 2 3g
If the function f is given by 1
2k x
f x e g x
, where k is a constant.
a. Find 1f , ' 1f , '' 1f
b. Show that the fourth derivative of f is
14 16 '''' 2k x
k e g x
a. 1 6f , ' 1 4f k , 2'' 1 12f k
b. 1
' 2 ' 2k x
f x ke g x
2 1'' 4 '' 2
k xf x k e g x
3 1''' 8 ''' 2
k xf x k e g x
4 1'''' 16 '''' 2
k xf x k e g x
81
2.2 More with Implicit Differentiation
As we saw in section 1.3, we could take the derivative of functions that are not
explicitly solved for y. We called this “implicit differentiation”. Now that we
have some more rules, our problems can get a bit more complex. In fact, we can
start to take second derivatives (or more) using this technique as well.
Ex. 1 Find the dy
dx for the function 4ln 5 22y x x
4ln 5 22yd
x xdx
3120 22
dyx
y dx
320 22dy
y xdx
Note, we solved for dy
dx by simply multiplying both
sides of our equation by y.
Don’t forget, we need the Chain Rule any time we take the derivative of a y.
OBJECTIVES
Find Derivatives of functions and relations that are not explicitly solved for y.
Ex 2 Find dy
dx and
2
2
d y
dx for 4ln 5 22y x x
We already found that 320 22dy
y xdx
from example 1, so now we simply
need to take the derivative again.
320 22d dy
y xdx dx
82
2
2
2
360 20 22yd y dy
x xdx dx
But we know that 320 22dy
y xdx
, so we can substitute!
2
2
2
3 360 20 22 20 22yd y
x x y xdx
2
22
2
360 20 22y yd y
x xdx
Note that the correct second derivative should not have dy
dx anywhere in the
problem – it needs to be substituted back in for the problem to be considered
finished.
These problems can get tricky because you can use a bunch of the rules that we
have already learned all in one problem.
Ex 3 Find dy
dx for
2
sinsec
xxy y
y
2
sinsec
d xxy y
dx y
2
4
cos 2 sinsec tan
dyy x y xdy dydxx y y y
dx dxy
Of course, isolating dy
dx can require some algebraic machinations! But
remember the basics – you can clear a fraction by multiplying everything by
the common denominator (in this case, y2).
We know the derivative of dy
dx is
2
2
d y
dx, so that takes care of the
left side of the equation, but we also have to recognize the right
side is a product, so we need the Product Rule.
You might notice we have a product, a
quotient, and some trig functions.
83
4 5 2 cos 2 sin sec tandy dy dy
xy y y x y x y ydx dx dx
4 5 2sec tan 2 sin cosdy dy dy
y y xy y x y y xdx dx dx
5 2
4
cos
sec tan 2 sin
dy y y x
dx y y xy y x
And, yes, the answers can be quite messy looking – but what would you expect?
Keep in mind, the derivative is the slope or rate of change of a function or relation,
and when the initial relation is complicated, describing its slope is often extremely
complicated.
Just so you can see, here is the graph of 2
sinsec
xxy y
y in a limited window:
x
y
With a graph that looks like this, of course the equation that represents the
slope of it is going to be rather complicated!
84
Ex 4 Find dy
dx and
2
2
d y
dx for 34 5 tan 4x y
32 5 tan 4d
x ydx
2 212 4sec 4dy
x ydx
2
2
3
sec 4
dy x
dx y
2 23 cos 4d dy
x ydx dx
2
2 2 2 22
24 cos 4 sin 4 3 cos 4 6 cos 4d y
x y y x y x ydx
2
2 3 22
72 cos 4 sin 4 6 cos 4d y
x y y x ydx
2
22
6 cos 4 1 12 cos 4 sin 4d y
x y x y ydx
Remember, we do substitute the dy
dx back into the equation. The final step in the
above example is just factoring out the common terms and rearranging the
subtraction so it looks a little nicer.
And, of course, anything we could do with derivatives, we still can do. So if we
were asked to come up with the equation of a tangent line for one of these
problems, there is very little extra work involved.
At this point, we could either use the Quotient Rule, or
we could recognize that the reciprocal of secant is
cosine and use the Product Rule. The second is the
strategy I will use, because it is a little easier in this
case.
85
2.2 Homework
Find dy
dx for each of the following:
1. 2sec xy x y e
2. 2 2 5yx y e x
3. sin 16sin
xy x y
y
86
4. Find dy
dx and
2
2
d y
dx for 2 2 xx y e
5. Find the equations of the lines tangent and normal to the curve 2 25 3 3x xy y x through the point (1,2).
87
Answers: 2.2 Homework
Find dy
dx for each of the following:
1. 2sec xy x y e
2
2
sec tan
xdy xy e
dx y y x
2. 2 2 5yx y e x
2 5
2ydy x
dx ye
3. sin 16sin
xy x y
y
2
2 2
cos sin sin
16sin cos sin sin
dy y x y y
dx y x y x y
4. Find dy
dx and
2
2
d y
dx for 2 2 xx y e
2
2
xdy e x
dx y
2
2 2
22 2 2
4
xx x e x
y e e xd y y
dx y
or
222
2 3
2 2 2
4
x xy e e xd y
dx y
5. Find the equations of the lines tangent and normal to the curve 2 25 3 3x xy y x through the point (1,2).
Tangent Line: 2 1 1y x
Normal Line: 2 1 1y x
88
2.3: Derivatives of Inverse Functions
We already know something about inverse functions. Exponential and
Logarithmic functions are inverses of each other, as are Radicals and Powers.
Inverse Functions—Defn: Two functions wherein the domain of one serves as the
range of the other and vice versa.
--Means: Two functions which cancel (or undo) each other.
The definition gives us a way to find the inverse for any function, the symbol for
which is f 1. You just switch the x and y variables and isolate y.
Ex 1 Find f 1 if f x x 3
f x x y x 3 3
So, for f 1, x y 3
y x
f x
3
1 3
This is little more interesting with a Rational Function.
Ex 2 Find f 1 if x
f xx
1
2
x x
f x yx x
1 1
2 2
89
So, for f 1,
yx
y
y x y
x xy y
x y xy
x y x
xy
xx
f xx
1
1
2
2 1
2 1
2 1
2 1 1
2 1
12 1
1
Note that - f f x x1 . For Example, f 1
02
and - f
1 1 02
.
General inverses are not all that interesting. We are more interested in particular
inverse functions, like the Ln. Another particular kind of inverse function that
bears more study is the Trig Inverse Function. Interestingly, as with the Log
functions, the derivatives of these Transcendental Functions become Algebraic
Functions.
Inverse Trig Derivative Rules
-
-
-
sin
cos
tan
u
u
u
du D
dx u
du D
dx u
du D
dx u
1
2
1
2
12
1
1
1
1
1
1
-
-
-
csc
sec
cot
u
u
u
du D
dx u u
du D
dx u u
du D
dx u
1
2
1
2
12
1
1
1
1
1
1
90
OBJECTIVE
Find the derivatives of inverse trig functions.
Proof that sin
dx
dx x
1
2
1
1.
sin sin y x y x 1
sin
cos
cos
xD y x
dyy
dx
dy
dx y
1
1
But this derivative is not in terms of x, so we are not done. Consider the
right triangle that would yield this SOHCAHTOA relationship:
1-x 2
1x
y
Note that for sin y to equal x, x must be the opposite leg and the hypotenuse
is 1 (SOH). The Pythagorean Theorem gives us the adjacent leg. By CAH,
cos y x 21
91
Therefore,
sin
cos
dx
dx y
x
1
2
1
1
1
Note that in the proof above, when we wanted to take the derivative of sin y, we
had to use The Chain Rule (since the y is a function other than x, and the derivative
of y is dy
dx.
Ex 4 - tan
dx
dx
1 43
-
tan d
x xdx
x
x
x
1 4 32
4
3
8
13 12
3 1
12
9 1
Ex 5 - sec
dx
dx
1 2
- sec
dx x
dxx x
x
x x
x x
1 2
22 2
22 2
4
12
1
2
1
2
1
92
2.3 Homework Set A
Find the derivative of the function. Simplify where possible.
1. 1sin xy e 2. 1tany x
3. 1sin 2 1y x 4. 2 1( ) 1 tan ( )H x x x
5. 1 2cos 1y x x x 6. 2( ) arctanxf x e x x
7. 1sin 2y x 8. 21csc 1y x
93
9. 1 11cot tany x
x
10. 21cos 1y x x x
11. 1sec
0x
y for xx
12. 2 1ln 4 tan2
xy x x
2.3 Homework Set B
1. 1 3cos zy e 2. 1 2tan 1y x
94
3. 1 214sin 4
2y x x x
4. 1 1cos
1
xy
x
5. 1 1sec 4 csc 4y x x 6. 1 2 2( ) sint
f t c c tc
7. 2( ) arccosf x x x 8. 1( ) ln tan 5f x x
95
9. 1 1sin 5 cos 5g w w w 10. 1 2( ) sec 9f t t
11. 2 1ln 1 coty u u u 12. 2
1 2tan
1
x
x
ey
e
96
Answers: 2.3 Homework Set A
1. 1sin xy e 2. 1tany x
21
x
x
dy e
dx e
312 2
1
2
dy
dxx x
3. 1sin 2 1y x 4. 2 1( ) 1 tan ( )H x x x
1
22
1dy
dxx x
1'( ) 1 2 tan ( )H x x x
5. 1 2cos 1y x x x 6. 2( ) arctanxf x e x x
1cosdy
dxx
2
2'( ) 2 arctan
1x x
f x e x xx
7. 1sin 2y x 8. 21csc 1y x
2
2
1 2
dy
dx x
2 2
2
1 2
dy x
dx x x x
9. 1 11cot tany x
x
10. 21cos 1y x x x
0dy
dx
2
2
3
1
dy x
dx x
11. 1sec
0x
y for xx
12. 2 1ln 4 tan2
xy x x
2
2 2
11
1
1 sec x
x
dy x
dx x
1tan
2
dy x
dx
97
2.3 Homework Set B
1. 1 3cos zy e 2. 1 2tan 1y x
3
6
3
1
z
z
dy e
dx e
2
1
1
dy
dx x x
3. 1 214sin 4
2y x x x
4. 1 1cos
1
xy
x
2
2
12 2
4
dy x
dx x
1
1
dy
dx x x
5. 1 1sec 4 csc 4y x x 6. 1 2 2( ) sint
f t c c tc
0dy
dx
2
2 2'( )
c tf t
c t
7. 2( ) arccosf x x x 8. 1( ) ln tan 5f x x
2
2'( ) 2 arccos
1
xf x x x
x
2 1
5'( )
1 25 tan 5f x
x x
9. 1 1sin 5 cos 5g w w w 10. 1 2( ) sec 9f t t
' 0g w 2 2
'( )9 8
tf t
t t
11. 2 1ln 1 coty u u u 12. 2
1 2tan
1
x
x
ey
e
213
cot1
dy uu
du u
2
2
1
x
x
dy e
dx e
98
2.4: Local Linearity and Approximations
Before calculators, one of the most valuable uses of the derivative was to find
approximate function values from a tangent line. Since the tangent line only shares
one point on the function, y-values on the line are very close to y-values on the
function. This idea is called local linearity—near the point of tangency, the
function curve appears to be a line. This can be easily demonstrated with the
graphing calculator by zooming in on the point of tangency. Consider the graphs
of .y x 425 and its tangent line at x 1 , .y x 75 .
x
y
x
The closer you zoom in, the more the line and the curve become one. The y-values
on the line are good approximations of the y-values on the curve. For a good
animation of this concept, see
http://www.ima.umn.edu/~arnold/tangent/tangent.mpg
Since it is easier to find the y-value of a line arithmetically than for other
functions—especially transcendental functions—the tangent line
approximation is useful if you have no calculator.
OBJECTIVES
Use the equation of a tangent line to approximate function values.
99
Ex 1 Find the tangent line equation to ( )f x x x x 4 3 22 1 at x = –1 and use it
to approximate value of ( . )f 0 9 .
The slope of the tangent line will be '( )f 1
'( )
'( )
f x x x x
f
3 24 3 4
1 3
[Note that we could have gotten this more easily with the nDeriv function on
our calculator.]
( )f 1 1, so the tangent line will be
( )
or
y x
y x
1 3 1
3 2
While we can find the exact value of ( . )f 0 9 with a calculator, we can get a
quick approximation from the tangent line. If x = –0.9 on the tangent line,
then
0.9 0.9 3 0.9 2 .7f y
This last example is somewhat trite in that we could have just plugged –0.9 into
( )f x x x x 4 3 22 1 and figured out the exact value even without a calculator.
It would have been a pain, but a person could actually do it by hand because of the
operations involved (there is only basic arithmetic involved). Consider the next
example, though.
100
Ex 2 Find the tangent line equation to xf x e 2 at x 0 and use it to
approximate value of .e0 2 .
Without a calculator, we could not find the exact value of .e0 2 . In fact, even
the calculator only gives an approximate value.
' xf x e 22 and ' xf e 20 2 2
f e 00 1
So the tangent line equation is y x 1 2 0 or y x 2 1
In order to find .e0 2 , we are looking for .f 0 1 . Therefore, we plug in
x = 0.1
. . .e 0 2 2 0 1 1 1 2
Note that the value that you get from a calculator for .e0 2 is 1.221403…
Our approximation of 1.2 seems very reasonable.
101
2.4 Homework Set A
1. Find the equation of the tangent line to 5( ) 5 1 at 2f x x x x and use it
to get an approximate value of 1.9f .
2. Find the tangent line equation to 2
5( ) at 1F x x x
x and use it to get an
approximate value of 1.1F .
3. Find all points on the graph of 22sin siny x x where the tangent line is
horizontal.
102
4. Find the equation of the tangent line at x = 2 for 2ln 3g x x . Use this
to approximate 2.1g .
5. Find the equation of the line tangent to 25 2 tan 1g x x x when
1x . Use this tangent line to find an approximation for 1.1g .
6. Use a tangent line to find the approximate value of .02f if
sin2 x xf x e x . Then use your calculator to find an actual value for .02f .
103
7. Find the equation of the tangent line for 2ln sin2
f x x x
for x = 1.
Plug in x = 1.1 into both the original function and the tangent line. Explain why
the values are so similar.
104
Answers: 2.4 Homework Set A
1. Find the equation of the tangent line to 5( ) 5 1 at 2f x x x x and use it
to get an approximate value of 1.9f .
21 75 2y x 1.9 13.5f
2. Find the tangent line equation to 2
5( ) at 1F x x x
x and use it to get an
approximate value of 1.1F .
21
4 12
y x 1.1 2.95F
3. Find all points on the graph of 22sin siny x x where the tangent line is
horizontal.
2 ,3 , 2 ,12 2
n n
4. Find the equation of the tangent line at x = 2 for 2ln 3g x x . Use this
to approximate 2.1g .
4 2y x 2.1 .4g
5. Find the equation of the line tangent to 25 2 tan 1g x x x when
1x . Use this tangent line to find an approximation for 1.1g .
7 4 1y x 1.1 7.4g
6. Use a tangent line to find the approximate value of .02f if
sin2 x xf x e x . Then use your calculator to find an actual value for .02f .
.02 2.04; .02 2.021...f f
105
7. Find the equation of the tangent line for 2ln sin2
f x x x
for x = 1.
Plug in x = 1.1 into both the original function and the tangent line. Explain why
the values are so similar.
1y x 1.1 .1; 1.1 .090f f
The results are so similar because of local linearity. The tangent line is
extremely close to the curve near the point of tangency, so values of y for the
tangent line serve as good approximations for values of f near the point of
tangency.
106
Chapter 2 Practice Test
1. Find 'f x , ''f x , and '"f x if ln cosf x x .
2. Find the equations of the lines tangent and normal to 1 2tan 3y x
when x = 2
3. Use a tangent line to find the approximate value of 0.01f if
21 cos2 xf x e x . Then use your calculator to find an actual value
for 0.01f .
107
4. Demonstrate that if 4tan 6y x , that 2
2
2288sec 6 tan 6
d y
dxx x .
5. Find dy
dx for the relation 2 sec 5yx y e .
6. Find the equation of the tangent line passing through the point (2,0) for the
relation 2 sec 5yx y e . Use it to approximate the value of y when x = 2.1
108
7. Find the derivative of each of the following functions:
a. 1sin 1y x b. 2 15 tanh x x x x
8. Find dy
dx and
2
2
d y
dx for the relation 2 sin 15x y x
109
Answers: Chapter 2 Practice Test
1. Find 'f x , ''f x , and '"f x if ln cosf x x .
' tanf x x
2'' secf x x
2''' 2sec tanf x x x
2. Find the equations of the lines tangent and normal to 1 2tan 3y x
when x = 2
Tangent: 2 24
y x
Normal: 1
24 2
y x
3. Use a tangent line to find the approximate value of 0.01f if
1 cos2 4xf x e x . Then use your calculator to find an actual value
for 0.01f .
0.01 0.01 1.96f y
0.01 1.9601...f
4. Demonstrate that if 4tan 6y x , that 22
2288sec 6 tan 6
d yx x
dx .
224sec 6dy
xdx
2
248sec 6 sec 6 tan 6 6
d yx x x
dx
22
2288sec 6 tan 6
d yx x
dx
5. Find dy
dx for the relation 2 sec 5yx y e .
2
2 sec
sec tan y
dy x y
dx x y y e
110
6. Find the equation of the tangent line passing through the point (2,0) for the
relation 2 sec 5yx y e . Use it to approximate the value of y when x = 2.1
Tangent line: 4 8y x
2.1 0.4y
7. Find the derivative of each of the following functions:
a. 1sin 1y x b. 2 15 tanh x x x x
1
2 1
dy
dx x x
1
2' 10 tan
1
xh x x x
x
8. Find dy
dx and
2
2
d y
dx for the relation 2 sin 15x y x
15 2
cos
dy x
dx y
22
2
2cos 15 2 tan
cos
y x yd y
ydx